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Question: A particle of sass 20g is released with an initial velocity \[5{\text{ m/s}}\] along the curve from ...

A particle of sass 20g is released with an initial velocity 5 m/s5{\text{ m/s}} along the curve from point A, as shown in the figure. The point A is at length h from B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum O will be: (Take g=10 ms210{\text{ m}}{{\text{s}}^{ - 2}})

A) 2 kg - m2s12{\text{ kg - }}{{\text{m}}^2}{{\text{s}}^{ - 1}}
B) 8 kg - m2s18{\text{ kg - }}{{\text{m}}^2}{{\text{s}}^{ - 1}}
C) 3 kg - m2s13{\text{ kg - }}{{\text{m}}^2}{{\text{s}}^{ - 1}}
D) 6 kg - m2s16{\text{ kg - }}{{\text{m}}^2}{{\text{s}}^{ - 1}}

Explanation

Solution

In this question, we need to determine the angular momentum of particle B when it reaches at point B. For this we will use the relation as L=r×p\vec L = \vec r \times \vec p, where
L\vec L is the angular momentum of a moving particle about a point
r\vec r is the length of perpendicular on the line of motion
p\vec p is the component of momentum along the perpendicular to r

Complete step by step answer:
Mass of the particle m=20gm = 20g
The velocity of the particle at point a is Va=5 m/s{V_a} = 5{\text{ m/s}}
Since the velocity of the particle will be different for two points, hence we apply work-energy theorem from point A to B
Wf=KE(i){W_f} = \vartriangle KE - - (i)
For two points equation (i) can be written as
mgh=12mvB212mvA2(ii)mgh = \dfrac{1}{2}mv_B^2 - \dfrac{1}{2}mv_A^2 - - (ii)
Equation (ii) can be further written as

mgh=12m(vB2vA2) 2gh=vB2vA2  mgh = \dfrac{1}{2}m\left( {v_B^2 - v_A^2} \right) \\\ 2gh = v_B^2 - v_A^2 \\\

Now by substitute the values of vA{v_A}andhh, we get

2gh=vB2vA2 2×10×10=vB2(5)2 vB225=200 vB2=225 vB=15 m/s  2gh = v_B^2 - v_A^2 \\\ \Rightarrow 2 \times 10 \times 10 = v_B^2 - {\left( 5 \right)^2} \\\ \Rightarrow v_B^2 - 25 = 200 \\\ \Rightarrow v_B^2 = 225 \\\ \Rightarrow {v_B} = 15{\text{ m/s}} \\\

Hence the angular momentum about O will be

L=mvrOB =151000×15×20 =6 kg - m2s1  L = mv{r_{OB}} \\\ = \dfrac{{15}}{{1000}} \times 15 \times 20 \\\ = 6{\text{ kg - }}{{\text{m}}^2}{{\text{s}}^{ - 1}} \\\

Option D is correct.

Note: Angular momentum is the same as linear momentum, but it is in rotational form. Angular momentum is a vector quantity, and it is the product of a body's rotational inertia and rotational velocity about a particular axis.